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Chapter 6: Problem 33

Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of \(1 .\) In each case, draw a graph, then find the probability of the given bone density test scores. If using technology instead of Table A-2, round answers to four decimal places. $$ \text { Less than } 4.55 $$

Short Answer

Expert verified
0.9999

Step by step solution

01

Understand the Problem

The problem requires finding the probability that a bone density test score is less than 4.55, given that the scores are normally distributed with a mean of 0 and a standard deviation of 1.
02

Draw the Standard Normal Distribution

Draw the standard normal distribution curve (a bell curve) with a mean (渭) of 0 and a standard deviation (蟽) of 1. Mark the value 4.55 on the horizontal axis, far to the right of the mean.
03

Convert the Test Score to a Z-Score (Optional)

Since the mean and standard deviation are already given for a standard normal distribution (mean = 0, standard deviation = 1), 4.55 is already the z-score.
04

Find the Probability

Using a Z-table or technology (like a calculator or statistical software), find the probability for a z-score less than 4.55. For z = 4.55, the corresponding probability (area under the curve to the left of 4.55) is approximately 0.9999968.
05

Confirm and Round the Probability

Since the problem specifies rounding the answer to four decimal places, round 0.9999968 to 0.9999.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Bone Density Test Scores
Bone density tests measure the strength and density of bones. These scores help in diagnosing osteoporosis and assessing fracture risks.
The scores are often compared to a reference population. This allows us to determine how a patient's bone density compares to what is expected.
When we say that the scores are 'normally distributed', we mean that they follow a specific statistical pattern. This pattern is shaped like a bell curve, where most scores cluster around the average value and fewer scores occur as they move further from the average.
Standard Normal Distribution
A standard normal distribution is a specific kind of normal distribution with a mean of 0 and a standard deviation of 1. It's a useful tool in statistics because it allows us to use tables and formulas to find probabilities.
The graph of a standard normal distribution is symmetrically bell-shaped, centered at zero.
When data follows this distribution, it means:
  • 68% of the data falls within 1 standard deviation from the mean (between -1 and 1).
  • 95% of the data falls within 2 standard deviations from the mean (between -2 and 2).
  • 99.7% of the data falls within 3 standard deviations from the mean (between -3 and 3).
Z-Score Calculations
A z-score tells us how many standard deviations a data point is from the mean.
The formula for calculating a z-score is:
\( z = \frac{(X - \mu)}{\sigma} \)
where:
  • \X \ is the raw score.
  • \mu \ is the mean of the distribution.
  • \sigma \ is the standard deviation of the distribution.
In our exercise, however, the mean is 0 and the standard deviation is 1. This means the given score (4.55) is already a z-score. Hence, the conversion isn鈥檛 necessary in this specific case.
Probability Using Normal Distribution
To find the probability of a z-score less than a given value, you can use a Z-table or technology.
For a standard normal distribution, the Z-table provides the cumulative probability for any z-score. This cumulative probability represents the area under the curve to the left of that z-score.
In our problem, we needed to find the probability for z = 4.55. According to the Z-table or software, the probability (area under the curve to the left of 4.55) is approximately 0.9999968.
Since the exercise asks us to round this value to four decimal places, we get 0.9999. This means there is a 99.99% chance that a randomly selected score will be less than 4.55, based on the normal distribution.

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Most popular questions from this chapter

Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of \(1 .\) In each case, draw a graph, then find the probability of the given bone density test scores. If using technology instead of Table A-2, round answers to four decimal places. $$ \text { Between }-2.00 \text { and } 2.00 $$

Use the data in the table below for sitting adult males and females (based on anthropometric survey data from Gordon, Churchill, et al.). These data are used often in the design of different seats, including aircraft seats, train seats, theater seats, and classroom seats. (Hint: Draw a graph in each case.) $$ \begin{aligned} &\text { Sitting Back-to-Knee Length (inches) }\\\ &\begin{array}{l|c|c|c} \hline & \text { Mean } & \text { St. Dev. } & \text { Distribution } \\ \hline \text { Males } & 23.5 \text { in. } & 1.1 \text { in. } & \text { Normal } \\ \hline \text { Females } & 22.7 \text { in. } & 1.0 \text { in. } & \text { Normal } \\ \hline \end{array} \end{aligned} $$ Instead of using \(0.05\) for identifying significant values, use the criteria th?t a value \(x\) is significantly high if \(P(x\) or greater \() \leq 0.01\) and a value is significantly low if \(P(x\) or less \() \leq 0.01\). Find the back-to- knee lengths for males, separating significant values from those that are not significant. Using these criteria, is a male back-to-knee length of 26 in. significantly high?

Do the following: If the requirements of \(n p \geq 5\) and \(n q \geq 5\) are both satisfied, estimate the indicated probability by using the normal distribution as an approximation to the binomial distribution; if \(n p<5\) or \(n q<5\), then state that the normal approximation should not be used. With \(n=20\) guesses and \(p=0.2\) for a correct answer, find \(P(\) at least 6 correct answers \()\).

Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of \(1 .\) In each case, draw a graph, then find the probability of the given bone density test scores. If using technology instead of Table A-2, round answers to four decimal places. $$ \text { Greater than } 0.25 $$

Annual incomes are known to have a distribution that is skewed to the right instead of being normally distributed. Assume that we collect a large \((n>30)\) random sample of annual incomes. Can the distribution of incomes in that sample be approximated by a normal distribution because the sample is large? Why or why not?
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